# How to prove that the $\angle CED$ of the triangle below is equal to $\frac{1}{2} \angle\alpha\;?$

Here is the whole problem:

In the triangle $$ABC$$, it is known that $$AC > AB$$, and the angle at the vertex $$A$$ is equal to $$\alpha$$. On the side $$AC$$, point $$M$$ is marked so that $$AB=MC$$. Point $$E$$ is the midpoint of the segment $$AM$$, point $$D$$ is the midpoint of the segment $$BC$$. Find the angle $$CED$$.

The textbook answer to the problem is that $$\angle CED = \dfrac{\alpha}{2}$$. The tip on how get to it is to mark a point $$K$$ on the $$AB$$ such, that $$AK = KB$$, draw the midline $$KD$$ of the triangle $$ABC$$ and then prove that the $$KDE$$ triangle is isosceles.

At that point I've tried everything(even chapgpt, which, as it turned out, is bad at math), the best I've gotten so far is if I mark a midpoint F on the $$CM$$, then I get parallelogram $$EKDF$$ $$\bigg(KD = \dfrac{1}{2}AC = EF$$, and $$KD\parallel AC\bigg)$$. And triangles $$KDE$$ and $$FED$$ are congruent, but it's not even close to what I need to prove. How would you do that ?

Here is the final figure:

Please, take into account that this is a problem from an $$\mathbf{8^{th}}$$ grade math textbook, the topic is "Midline of a triangle", which means I'm not allowed to use any angle functions or similar features that were not introduced yet in the curicullumn.

EDIT: The textbook's hint seems to be wrong, thanks @Vasili for the solution. The actual triangle that needs to be proven to be isosceles is another one, not the $$KDE$$.

• Could you provide more details like what you have studied under the topic mentioned.
– KKT
Commented May 2, 2023 at 14:13
• @KKT, should I list all the material I've gone through ? :) I think I gave sufficient amount of information to the particular problem ? Commented May 2, 2023 at 14:31
• The hint “Prove that the $KDE$ triangle is isosceles” seems to be a mistake. The key, as @Vasili shows, is rather that $\triangle NDE$ is isosceles. Commented May 2, 2023 at 22:21
• @EdwardPorcella, yeah, that's what I figured when I saw @Vasili solution. Commented May 2, 2023 at 22:40

Here is how I arrived to the solution but also without completely using the hint.
Draw $$BM$$ and $$NE$$. $$NE$$ is a midline in $$\triangle ABM$$ so $$AKNE$$ is a paralellogram, $$NE=AK$$. $$ND$$ is a midline in $$\triangle BMC$$ so $$ND=\frac{MC}{2}=AK=NE \implies \triangle END$$ is isosceles.

• I have no time to review it completely right now, but without BM I understand the proof. I'll check it up later. Thanks for the answer. Commented May 2, 2023 at 15:26
• $BM$ is needed to show that $ND=\frac{MC}{2}$ Commented May 2, 2023 at 15:29
• I just checked your solution, it's correct, thank you very much! I guess the textbook's hint is just wrong But I'd like to add that BM seems to be redundant here. Simply mark point $N$ so that $KN = AE$ => $ND = (KD=EF) - (KN=EM) = MF$. And yeah, from then on we simply prove that $\triangle END$ is isosceles. Thanks again! Commented May 2, 2023 at 23:02

As $$\triangle BKD \sim \triangle BAC$$, $$|KD|=L_1+L_3$$.

Translating D to the left by $$L_3$$, and moving E correspondingly (i.e. to A), maintains $$\angle DEF$$ and creates a parallelogram with all sides equal to $$L_1$$, of which the translated line $$D'E'$$ is a diagonal bisector.

• what is the "right line" and the "translated line"?
– D S
Commented May 2, 2023 at 13:02
• Thanks for your answer, but there were no similarity of triangles yet in the curicullumn. I don't really understand what "L1", "L3", "moving E", "translated line" mean. I provided a hint from the textbook, would be nice to see a solution in accordance to that hint. Commented May 2, 2023 at 13:06
• @DS; I haven't. The orange line completes the parallelogram from $AKD'$. $|KD'|=L_1$, as does $|AK|$.